Đáp án:
$AM=2; HM=\sqrt{3}; BH=2 \pm \sqrt{3}; CH=4-BH=2 \mp \sqrt{3};\\ AB=\sqrt{8 \pm 4\sqrt{3}}; AC=\sqrt{8 \mp 4\sqrt{3}} $
Giải thích các bước giải:
$\Delta ABC$ vuông tại $A$, đường cao $AH$
$\Rightarrow \left\{\begin{array}{l} AB^2+AC^2=BC^2\\ AB.AC=AH.BC\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB^2+AC^2=16\\ AB.AC=4\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB^2+AC^2=16\\ AC=\dfrac{4}{AB}\end{array} \right.\\ \Leftrightarrow \left\{\begin{array}{l} AB^2+\dfrac{16}{AB^2}=16\\ AC=\dfrac{4}{AB}\end{array} \right.\\ \Leftrightarrow \left[\begin{array}{l} \left\{\begin{array}{l} AB=\sqrt{8+4\sqrt{3}}\\ AC=\sqrt{8-4\sqrt{3}}\end{array} \right.\\ \left\{\begin{array}{l} AB=\sqrt{8-4\sqrt{3}}\\ AC=\sqrt{8+4\sqrt{3}}\end{array}\right. \end{array} \right.$
Trung tuyến $AM=\dfrac{1}{2}BC=2$
$HM=\sqrt{AM^2-AH^2}=\sqrt{3}\\ BH=\sqrt{AB^2-AH^2}=2 \pm \sqrt{3}\\ CH=4-BH=2 \mp \sqrt{3}$
